Henderson Hasselbalch Equation

Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach everything you need to know about Henderson Hasselbalch Equation of Acids and Bases. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions. We can identify buffer solutions based on the presence of a conjugate acid-base pair. Consider a buffer with a pKa of 4.8 and think about the buffer system under 5 different pH conditions. Based on our equation, if we know the pH and pKa of a buffer system, we can calculate the proportion of base concentration to acid concentration. We will develop a picture of each situation with a chart that describes the proportion of acid and base that has been protonated and deprotonated.
Case 1. pH = 4.8; pKa = 4.8. In this instance, the pH and pKa of the buffer system are equal. Substituting these values into the H-H equation, we find that log([A-]/[HA]) must be equal to 0. Our goal is to find the ratio of base to acid, so we want to remove the log surrounding the term [A-]/[HA]. In order to remove a standard log, we take both sides to the power of 10. 100 = 1, and 10log[A]/[HA] = [A-]/[HA]. Therefore, when pH = pKa, the ratio of conjugate base to acid is 1-to-1. In other words, 50% of our solution is in the deprotonated form, and 50% is in the protonated form.
Case 2. pH = 3.8; pKa = 4.8. In this instance, pH is less than the pKa of the buffer system. Substituting these values into the H-H equation, we find that must be equal to -1. Our goal is to find the ratio of base to acid, so we want to remove the log surrounding the term [A-]/[HA]. In order to remove a standard log, we take both sides to the power of 10. 10-1 = 1/10, and 10log[A]/[HA] = [A-]/[HA]. Therefore, when pH is 1 less than pKa, the ratio of conjugate base to acid is 1 to 10 In other words, 10% of our solution is in the deprotonated form, and 90% is in the protonated form.
Case 3. pH = 2.8; pKa = 4.8. Substituting these values into the H-H equation, we find that log([A-]/[HA]) must be equal to -2. In order to remove the log surrounding the ratio of products to reactants, we take both sides to the power of 10. 10-2 = 1/100, and 10log[A]/[HA] = [A-]/[HA]. Therefore, when pH is 2 less than pKa, the ratio of conjugate base to acid is 1 to 100 In other words, 1% of our solution is in the deprotonated form, and 99% is in the protonated form.
Case 4. pH = 1.8; pKa = 4.8. Substituting these values into the H-H equation, we find that log([A-]/[HA]) must be equal to -3. In order to remove the log surrounding the ratio of products to reactants, we take both sides to the power of 10. 10-3 = 1/1000, and 10log[A]/[HA] = [A-]/[HA]. Therefore, when pH is 3 less than pKa, the ratio of conjugate base to acid is 1 to 1000 In other words, 0.1% of our solution is in the deprotonated form, and 99.9% is in the protonated form.
Case 5. pH = 6.8; pKa = 4.8. Here, we consider a situation where pH greater than pKa. Substituting these values into the H-H equation, we find that log([A-]/[HA]) must be equal to 2. In order to remove the log surrounding the ratio of products to reactants, we take both sides to the power of 10. 102 = 100/1, and 10log[A]/[HA] = ([A-]/[HA]) Therefore, when pH is 2 greater than pKa, the ratio of conjugate base to acid is 100 to 1 In other words, 99% of our solution is in the deprotonated form, and 1% is in the protonated form.
The takeaway is that for any change in pH, the ratio of products to reactants changes by a factor of 10. When pH is greater than the buffer’s pKa, the system strongly favors the basic, deprotonated form of the conjugate acid-base pair. In this case, the abundance of base means that the system will be really successful at minimizing changes in pH in response to any acid added to the system because the excess of base will be excellent at neutralizing new acid. In converse, when the pH is less than the pKa, the system has a much greater concentration of conjugate acid, which will be excellent at neutralizing any excess base added to the solution. In both cases (pH less than pKa and pH greater than pKa), the buffer is okay at neutralizing something, but not excellent at neutralizing both acids and bases. However, when pH = pKa, we note that there is an abundance of both members of the conjugate acid-base pair. Therefore, when the system’s pH is equal to the pKa of our chosen buffer, the buffer will perform well at neutralizing any excess acid or base added.
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[@justinpalmer8608]

Hello! Thank you for the wonderful videos, I have watched every single video on this playlist up to this point and they have cleared a lot of concepts up! I did have a quick question though, I am a little confused as to why the [A-] would go down as pH goes down. I thought that with acid strength, more protons would be released, which would mean that the [A-] would go up and [HA] would go down (because more protons are being released from that acid).
It definitely makes sense when deconstructing the equation and plugging in those values, like for example, to lower pH by 1 it would have to be log(1/10), I guess conceptually I am just struggling to wrap my brain around this.
Thank you for the wonderful videos they have been a huge help with my studies!! :)

[@StarSunYT]

As more acid is added, there is more H+ in solution, decreasing the pH. That extra H+ would combine with the A- in solution increasing the HA concentration while decreasing the A- concentration in order to make that HA.

[@justinpalmer8608]

@@StarSunYT thank you!